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| [[File:Combustion reaction of methane.jpg|thumb|400px|As seen from the equation {{chem|CH|4}} + 2{{chem|O|2}} → {{chem|CO|2}} + 2{{chem|H|2|O}}, one molecule of [[methane]] reacts with two molecules of [[oxygen]] gas to yield one molecule of [[carbon dioxide]] and two molecules [[Properties of water|water]]. Stoichiometry describes these types of quantitative relationships, as well being used to determine the amount of products/reactants that are produced/needed in a given reaction]]
| | == は「これで、行くと長老たちは話す == |
| '''Stoichiometry''' {{IPAc-en|ˌ|s|t|ɔɪ|k|i|ˈ|ɒ|m|ɨ|t|r|i}} is a branch of [[chemistry]] that deals with the relative quantities of [[reactant]]s and [[Product (chemistry)|products]] in [[chemical reaction]]s. In a balanced chemical reaction, the relations among quantities of reactants and products typically form a ratio of positive integers. For example, in a reaction that forms [[ammonia]] (NH<sub>3</sub>), exactly one molecule of [[nitrogen]] gas (N<sub>2</sub>) reacts with three molecules of [[hydrogen]] gas (H<sub>2</sub>) to produce two molecules of NH<sub>3</sub>:
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| : {{chem|N|2}} + 3{{chem|H|2}} → 2{{chem|NH|3}}
| | その他本体は、私は彼らがすでに知っている、子供の体の事ではないです。 [http://www.ispsc.edu.ph/nav/japandi/casio-rakuten-3.html カシオ gps 時計] '<br><br>シャオヤン少し開いた口、直ちに中空笑い、口の音が発行されている間、頭に沿ってテイクに沿って取る、恥ずかしさの「色」を見て、良い瞬間の後、ちょうど慎重に言った： [http://www.nnyagdev.org/sitemap.xml http://www.nnyagdev.org/sitemap.xml] '？。彼らはナットを行くこと」<br><br>はかすかな、、シャオヤン斜め神 '色'メデューサの心を少し笑いを見た頬が、それはまだ寒いです。「家族のルールによると、あなたがワンヘビにかま本体の影響を受けます。 [http://www.ispsc.edu.ph/nav/japandi/casio-rakuten-10.html カシオ腕時計 価格] '<br><br>その何ワンヘビにかま本体を聞き、シャオヤンは厳しく、乾いた含み笑いを身震いされています。「私たちは、私は今私が何をするための炎症リーグのチーフ、咳、午前、これらの事を使用するように求めていないものを、言って良い何かを持っているこれらのことは、確かにガマとテラン帝国は、再びひずみヘビになりますように素晴らしいではない、誰も。 [http://www.ispsc.edu.ph/nav/japandi/casio-rakuten-11.html カシオ腕時計 g-shock] '<br><br>は「これで、行くと長老たちは話す [http://www.ispsc.edu.ph/nav/japandi/casio-rakuten-9.html カシオ 掛け時計]。「メデューサQingtai目、鈍いサウンドトラック。<br><br>シャオヤンクラッチ |
| | | 相关的主题文章： |
| This particular kind of stoichiometry - describing the quantitative relationships among substances as they participate in chemical reactions - is known as ''reaction stoichiometry''. In the example above, reaction stoichiometry describes the 1:3:2 ratio of molecules of nitrogen, hydrogen, and ammonia.
| | <ul> |
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| Stoichiometry can be used to determine quantities such as the amount of products (in mass, moles, volume, etc.) that can be produced with given reactants and percent [[Yield (chemistry)|yield]] (the percentage of the given reactant that is made into the product). Stoichiometry calculations can predict how elements and components diluted in a [[standard solution]] react in experimental conditions. Stoichiometry is founded on the [[law of conservation of mass]]: the mass of the reactants equals the mass of the products.
| | <li>[http://www.408yy.com/plus/feedback.php?aid=2061 http://www.408yy.com/plus/feedback.php?aid=2061]</li> |
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| ''Composition stoichiometry'' describes the quantitative (mass) relationships among elements in compounds. For example, composition stoichiometry describes the nitrogen to hydrogen ratio in the compound ammonia (NH<sub>3</sub>): 1 [[Mole (unit)|mol]] of ammonia consists of 1 mol of nitrogen and 3 mol of hydrogen. As the nitrogen atom is about 14 times heavier than the hydrogen atom, the mass ratio is 14:3, thus 17 kg of ammonia contains 14 kg of nitrogen and 3 kg of hydrogen.
| | <li>[http://www.fukufukutei.jp/cgi-bin/hikotin/bbs/honey.cgi http://www.fukufukutei.jp/cgi-bin/hikotin/bbs/honey.cgi]</li> |
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| A ''stoichiometric amount'' or ''stoichiometric ratio'' of a [[reagent]] is the optimum amount or ratio where, assuming that the reaction proceeds to completion:
| | <li>[http://guian.gzlcbbs.com/forum.php?mod=viewthread&tid=3413 http://guian.gzlcbbs.com/forum.php?mod=viewthread&tid=3413]</li> |
| # All of the reagent is consumed
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| # There is no deficiency of the reagent
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| # There is no excess of the reagent.
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| A non-stoichiometric mixture, wherein reactions have gone to completion, will have only the [[limiting reagent]] consumed completely.
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| While almost all reactions have integer-ratio stoichiometry in amount of matter units (moles, number of particles), some [[nonstoichiometric compound]]s that cannot be represented by a ratio of well-defined natural numbers are known. These materials, therefore, violate the [[law of definite proportions]] that forms the basis of stoichiometry along with the [[law of multiple proportions]].
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| ''Gas stoichiometry'' deals with reactions involving gases, where the gases are at a known temperature, pressure, and volume, and can be assumed to be [[ideal gas]]es. For gases, the volume ratio is ideally the same by the [[ideal gas law]], but the mass ratio of a single reaction has to be calculated from the [[molecular mass]]es of the reactants and products. In practice, due to the existence of [[isotope]]s, [[molar mass]]es are used instead when calculating the mass ratio.
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| ==Etymology==
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| The term ''stoichiometry'' was first used by [[Jeremias Benjamin Richter]] in 1792 when the first volume of Richter's ''Stoichiometry or the Art of Measuring the Chemical Elements'' was published. The term is derived from the [[Greek language|Greek]] words στοιχεῖον ''stoicheion'' "element" and μέτρον ''metron'' "measure". In [[patristic]] Greek, the word ''Stoichiometria'' was used by [[Patriarch Nicephorus I of Constantinople|Nicephorus]] to refer to the number of line counts of the [[Biblical canon|canonical]] [[New Testament]] and some of the [[Apocrypha]].
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| ==Definition==
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| Stoichiometry rests upon the very basic laws that help to understand it better, i.e., [[law of conservation of mass]], the [[law of definite proportions]] (i.e., the [[law of constant composition]]), and the [[law of multiple proportions]]. In general, chemical reactions combine in definite ratios of chemicals. Since chemical reactions can neither create nor destroy matter, nor [[nuclear transmutation|transmute]] one element into another, the amount of each element must be the same throughout the overall reaction. For example, the number of atoms of a given element X on the reactant side must equal the number of atoms of that element on the product side, whether or not all of those atoms are actually involved in a reaction.
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| Chemical reactions, as macroscopic unit operations, consist of simply a very large number of [[elementary reaction]]s, where a single molecule reacts with another molecule. As the reacting molecules (or moieties) consist of a definite set of atoms in an integer ratio, the ratio between reactants in a complete reaction is also in integer ratio. A reaction may consume more than one molecule, and the '''stoichiometric number''' counts this number, defined as positive for products (added) and negative for reactants (removed).<ref>[http://goldbook.iupac.org/S06025-plain.html goldbook.iupac.org]</ref>
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| Different elements have a different [[atomic mass]], and as collections of single atoms, molecules have a definite [[molar mass]], measured with the unit mole (6.02 × 10<sup>23</sup> individual molecules, [[Avogadro's constant]]). By definition, carbon-12 has a molar mass of 12 g/mol. Thus, to calculate the stoichiometry by mass, the number of molecules required for each reactant is expressed in moles and multiplied by the molar mass of each to give the mass of each reactant per mole of reaction. The mass ratios can be calculated by dividing each by the total in the whole reaction.
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| Elements in their natural state are mixtures of [[isotope]]s of differing mass, thus [[atomic mass]]es and thus molar masses are not exactly integers. For instance, instead of an exact 14:3 proportion, 17.04 kg of ammonia consists of 14.01 kg of nitrogen and 3 × 1.01 kg of hydrogen, because natural nitrogen includes a small amount of nitrogen-15, and natural hydrogen includes hydrogen-2 ([[deuterium]]).
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| == Converting grams to moles ==
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| Stoichiometry is not only used to balance chemical equations but also used in conversions, i.e., converting from grams to moles, or from grams to milliliters. For example, to find the number of moles in 2.00 g of NaCl, one would do the following:
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| :<math>\frac{2.00 \mbox{ g NaCl}}{58.44 \mbox{ g NaCl mol}^{-1}} = 0.034 \ \text{mol}</math>
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| In the above example, when written out in fraction form, the units of grams form a multiplicative identity, which is equivalent to one (g/g=1), with the resulting amount of moles (the unit that was needed), is shown in the following equation,
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| :<math>\left(\frac{2.00 \mbox{ g NaCl}}{1}\right)\left(\frac{1 \mbox{ mol NaCl}}{58.44 \mbox{ g NaCl}}\right) = 0.034\ \text{mol}</math>
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| ==Molar proportions ==
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| Stoichiometry is often used to balance chemical equations (reaction stoichiometry). For example, the two [[Diatomic molecule|diatomic]] gases, [[hydrogen]] and [[oxygen]], can combine to form a liquid, water, in an [[exothermic reaction]], as described by the following equation:
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| : 2{{chem|H|2}} + {{chem|O|2}} → 2{{chem|H|2|O}}
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| Reaction stoichiometry describes the 2:1:2 ratio of hydrogen, oxygen, and water molecules in the above equation.
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| The molar ratio allows for conversion between moles of one substance and moles of another. For example, in the reaction
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| 2{{chem|CH|3|OH}} + 3{{chem|O|2}} → 2{{chem|CO|2}} + 4{{chem|H|2|O}}
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| the number of moles of water that will be produced by the combustion of 0.27 moles of {{chem|CH|3|OH}} is obtained using the molar ratio between {{chem|CH|3|OH}} and {{chem|H|2|O}} of 2 to 4.
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| :<math>\left(\frac{0.27 \mbox{ mol }\mathrm{CH_3OH}}{1}\right)\left(\frac{4 \mbox{ mol }\mathrm{H_2O}}{2 \mbox{ mol } \mathrm{CH_3OH}}\right) = 0.54\ \text{mol}</math>
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| The term stoichiometry is also often used for the [[Mole (unit)|molar]] proportions of elements in stoichiometric compounds (composition stoichiometry). For example, the stoichiometry of hydrogen and oxygen in H<sub>2</sub>O is 2:1. In stoichiometric compounds, the molar proportions are whole numbers.
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| ==Determining amount of product==
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| Stoichiometry can also be used to find the quantity of a product yielded by a reaction. If a piece of solid [[copper]] (Cu) were added to an aqueous solution of [[silver nitrate]] (AgNO<sub>3</sub>), the [[silver]] (Ag) would be replaced in a [[single displacement reaction]] forming aqueous [[copper(II) nitrate]] (Cu(NO<sub>3</sub>)<sub>2</sub>) and solid silver. How much silver is produced if 16.00 grams of Cu is added to the solution of excess silver nitrate?
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| The following steps would be used:
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| :Step 1 - Write and Balance the Equation
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| :Step 2 - Mass to Mole: Convert g Cu to moles Cu
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| :Step 3 - Mole Ratio: Convert moles of Cu to moles of Ag produced
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| :Step 4 - Mole to Mass: Convert moles Ag to grams of Ag produced
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| The complete balanced equation would be:
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| : {{chem|Cu}} + 2{{chem|Ag|NO|3}} → {{chem|Cu|(NO|3|)|2}} + 2{{chem|Ag}}
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| For the mass to mole step, the amount of copper (16.00 g) would be converted to moles of copper by dividing the mass of copper by its [[molecular mass]]: 63.55 g/mol.
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| :<math>\left(\frac{16.00 \mbox{ g Cu}}{1}\right)\left(\frac{1 \mbox{ mol Cu}}{63.55 \mbox{ g Cu}}\right) = 0.2518\ \text{mol Cu}</math> | |
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| Now that the amount of Cu in moles (0.2518) is found, we can set up the mole ratio. This is found by looking at the coefficients in the balanced equation: Cu and Ag are in a 1:2 ratio.
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| :<math>\left(\frac{0.2518 \mbox{ mol Cu}}{1}\right)\left(\frac{2 \mbox{ mol Ag}}{1 \mbox{ mol Cu}}\right) = 0.5036\ \text{mol Ag}</math>
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| Now that the moles of Ag produced is known to be 0.5036 mol, we convert this amount to grams of Ag produced to come to the final answer:
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| :<math>\left(\frac{0.5036 \mbox{ mol Ag}}{1}\right)\left(\frac{107.87 \mbox{ g Ag}}{1 \mbox{ mol Ag}}\right) = 54.32 \ \text{g Ag}</math>
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| This set of calculations can be further condensed into a single step:
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| :<math>m_\mathrm{Ag} = \left(\frac{16.00 \mbox{ g }\mathrm{Cu}}{1}\right)\left(\frac{1 \mbox{ mol }\mathrm{Cu}}{63.55 \mbox{ g }\mathrm{Cu}}\right)\left(\frac{2 \mbox{ mol }\mathrm{Ag}}{1 \mbox{ mol }\mathrm{Cu}}\right)\left(\frac{107.87 \mbox{ g }\mathrm{Ag}}{1 \mbox{ mol Ag}}\right) = 54.32 \mbox{ g}</math>
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| ===Further examples===
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| For [[propane]] (C<sub>3</sub>H<sub>8</sub>) reacting with [[oxygen|oxygen gas]] (O<sub>2</sub>), the balanced chemical equation is:
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| : {{chem|C|3|H|8}} + 5{{chem|O|2}} → 3{{chem|CO|2}} + 4{{chem|H|2|O}}
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| The mass of water formed if 120 g of propane (C<sub>3</sub>H<sub>8</sub>) is burned in excess oxygen is then
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| :<math>m_\mathrm{H_2O} = \left(\frac{120. \mbox{ g }\mathrm{C_3H_8}}{1}\right)\left(\frac{1 \mbox{ mol }\mathrm{C_3H_8}}{44.09 \mbox{ g }\mathrm{C_3H_8}}\right)\left(\frac{4 \mbox{ mol }\mathrm{H_2O}}{1 \mbox{ mol }\mathrm{C_3H_8}}\right)\left(\frac{18.02 \mbox{ g }\mathrm{H_2O}}{1 \mbox{ mol }\mathrm{H_2O}}\right) = 196 \mbox{ g}</math>
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| ==Stoichiometric ratio==
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| Stoichiometry is also used to find the right amount of one [[reactant]] to "completely" react with the other reactant in a [[chemical reaction]] - that is, the stoichiometric amounts that would result in no leftover reactants when the reaction takes place. An example is shown below using the [[thermite reaction]],
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| : {{chem|Fe|2|O|3}} + 2{{chem|Al}} → {{chem|Al|2|O|3}} + 2{{chem|Fe}}
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| This equation shows that 1 mole of {{nowrap|[[iron(III) oxide]]}} and 2 moles of [[aluminum]] will produce 1 mole of [[aluminium oxide]] and 2 moles of [[iron]]. So, to completely react with 85.0 g of {{nowrap|iron(III) oxide}} (0.532 mol), 28.7 g (1.06 mol) of aluminium are needed.
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| :<math>m_\mathrm{Al} = \left(\frac{85.0 \mbox{ g }\mathrm{Fe_2O_3}}{1}\right)\left(\frac{1 \mbox{ mol }\mathrm{Fe_2 O_3}}{159.7 \mbox{ g }\mathrm{Fe_2 O_3}}\right)\left(\frac{2 \mbox{ mol Al}}{1 \mbox{ mol }\mathrm{Fe_2 O_3}}\right)\left(\frac{26.98 \mbox{ g Al}}{1 \mbox{ mol Al}}\right) = 28.7 \mbox{ g}</math>
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| ==Limiting reagent and percent yield==
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| {{Main|Limiting reagent|Yield (chemistry)}}
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| The limiting reagent is the reagent that limits the amount of product that can be formed and is completely consumed during the reaction. The excess reactant is the reactant that is left over once the reaction has stopped due to the limiting reactant.
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| Consider the equation of roasting [[lead(II) sulfide]] (PbS) in oxygen (O<sub>2</sub>) to produce [[lead(II) oxide]] (PbO) and [[sulfur dioxide]] (SO<sub>2</sub>):
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| : 2{{chem|PbS}} + 3{{chem|O|2}} → 2{{chem|PbO}} + 2{{chem|SO|2}}
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| To determine the theoretical yield of lead(II) oxide if 200.0 g of lead(II) sulfide and 200.0 grams of oxygen are heated in an open container:
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| :<math>m_\mathrm{PbO} = \left(\frac{200.0 \mbox{ g }\mathrm{PbS}}{1}\right)\left(\frac{1 \mbox{ mol }\mathrm{PbS}}{239.27 \mbox{ g }\mathrm{PbS}}\right)\left(\frac{2 \mbox{ mol }\mathrm{PbO}}{2 \mbox{ mol }\mathrm{PbS}}\right)\left(\frac{223.2 \mbox{ g }\mathrm{PbO}}{1 \mbox{ mol }\mathrm{PbO}}\right) = 186.6 \mbox{ g}</math>
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| :<math>m_\mathrm{PbO} = \left(\frac{200.0 \mbox{ g }\mathrm{O_2}}{1}\right)\left(\frac{1 \mbox{ mol }\mathrm{O_2}}{32.00 \mbox{ g }\mathrm{O_2}}\right)\left(\frac{2 \mbox{ mol }\mathrm{PbO}}{3 \mbox{ mol }\mathrm{O_2}}\right)\left(\frac{223.2 \mbox{ g }\mathrm{PbO}}{1 \mbox{ mol }\mathrm{PbO}}\right) = 930.0 \mbox{ g}</math>
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| Because a lesser amount of PbO is produced for the 200.0 g of PbS, it is clear that PbS is the limiting reagent.
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| In reality, the actual yield is not the same as the stoichiometrically-calculated theoretical yield. Percent yield, then, is expressed in the following equation:
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| <center><math>\mbox{percent yield} = \frac{\mbox{actual yield}}{\mbox{theoretical yield}} \times \!\, 100</math></center> | |
| If 170.0 g of lead(II) oxide is obtained, then the percent yield would be calculated as follows:
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| <center><math>\mbox{percent yield} = \frac{\mbox{170.0 g PbO}}{\mbox{186.6 g PbO}} \times \!\, 100 = 91.12\%</math></center>
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| ===Example===
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| Consider the following reaction, in which [[iron(III) chloride]] reacts with [[hydrogen sulfide]] to produce [[iron(III) sulfide]] and [[hydrogen chloride]]:
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| : 2{{chem|FeCl|3}} + 3{{chem|H|2|S}} → {{chem|Fe|2|S|3}} + 6{{chem|HCl}}
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| Suppose 90.0 g of FeCl<sub>3</sub> reacts with 52.0 g of H<sub>2</sub> S. To find the limiting reagent and the mass of HCl produced by the reaction, we could set up the following equations:
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| :<math>m_\mathrm{HCl} = \left(\frac{90.0 \mbox{ g }\mathrm{FeCl_3}}{1}\right)\left(\frac{1 \mbox{ mol }\mathrm{FeCl_3}}{162 \mbox{ g }\mathrm{FeCl_3}}\right)\left(\frac{6 \mbox{ mol }\mathrm{HCl}}{2 \mbox{ mol }\mathrm{FeCl_3}}\right)\left(\frac{36.5 \mbox{ g }\mathrm{HCl}}{1 \mbox{ mol }\mathrm{HCl}}\right) = 60.8 \mbox{ g}</math>
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| :<math>m_\mathrm{HCl} = \left(\frac{52.0 \mbox{ g }\mathrm{H_2S}}{1}\right)\left(\frac{1 \mbox{ mol }\mathrm{H_2S}}{34.1 \mbox{ g }\mathrm{H_2S}}\right)\left(\frac{6 \mbox{ mol }\mathrm{HCl}}{3 \mbox{ mol }\mathrm{H_2S}}\right)\left(\frac{36.5 \mbox{ g }\mathrm{HCl}}{1 \mbox{ mol }\mathrm{HCl}}\right) = 111 \mbox{ g}</math>
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| Thus, the limiting reagent is FeCl<sub>3</sub> and the amount of HCl produced is 60.8 g.
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| To find what mass of excess reagent (H<sub>2</sub>S) remains after the reaction, we would set up the calculation to find out how much H<sub>2</sub>S reacts completely with the 90.0 g FeCl<sub>3</sub>:
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| :<math>m_\mathrm{H_2S} = \left(\frac{90.0 \mbox{ g }\mathrm{FeCl_3}}{1}\right)\left(\frac{1 \mbox{ mol }\mathrm{FeCl_3}}{162 \mbox{ g }\mathrm{FeCl_3}}\right)\left(\frac{3 \mbox{ mol }\mathrm{H_2S}}{2 \mbox{ mol }\mathrm{FeCl_3}}\right)\left(\frac{34.1 \mbox{ g }\mathrm{H_2S}}{1 \mbox{ mol }\mathrm{H_2S}}\right) = {28.4 \mbox{ g }\mathrm{reacted}}</math>
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| By subtracting this amount from the original amount of H<sub>2</sub>S, we can come to the answer:
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| :<math>{52.0 \mbox{ g }\mathrm{H_2S}} - {28.4 \mbox{ g }\mathrm{H_2S}} = {23.6 \mbox{ g }\mathrm{H_2S}}</math> <math>{\mathrm{excess}}</math>
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| ==Different stoichiometries in competing reactions==
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| Often, more than one reaction is possible given the same starting materials. The reactions may differ in their stoichiometry. For example, the [[methylation]] of [[benzene]] (<math>\mathrm{C_6H_6}</math>), through a [[Friedel-Crafts reaction]] using <math>\mathrm{AlCl_3}</math> as a catalyst, may produce singly methylated <math>(\mathrm{C_6H_5CH_3})</math>, doubly methylated <math>(\mathrm{C_6H_4(CH_3)_2})</math>, or still more highly methylated <math>(\mathrm{C_6H}_{6-n}(\mathrm{CH_3})_n)</math> products, as shown in the following example,
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| : <math>\mathrm{C_6H_6 + CH_3Cl \rightarrow C_6H_5CH_3 + HCl}\,</math> | |
| : <math>\mathrm{C_6H_6 + \,2\ CH_3Cl \rightarrow C_6H_4(CH_3)_2 + 2HCl}\,</math>
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| : <math>\mathrm{C_6H_6} + \,n\ \mathrm{CH_3Cl} \rightarrow \mathrm{C_6H}_{6-n}(\mathrm{CH_3})_n + n\,\mathrm{HCl}\,</math>
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| In this example, which reaction takes place is controlled in part by the relative [[concentration]]s of the reactants.
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| ==Stoichiometric coefficient==
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| In lay terms, the ''stoichiometric coefficient'' (or ''stoichiometric number'' in the IUPAC nomenclature<ref>IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"). Compiled by A. D. McNaught and A. Wilkinson. Blackwell Scientific Publications, Oxford (1997). XML on-line corrected version: http://goldbook.iupac.org (2006-) created by M. Nic, J. Jirat, B. Kosata; updates compiled by A. Jenkins. ISBN 0-9678550-9-8. {{doi|10.1351/goldbook}}. Entry: [http://goldbook.iupac.org/S06025.html "stoichiometric number"].</ref>) of any given component is the number of molecules that participate in the reaction as written.<br />
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| For example, in the reaction CH<sub>4</sub> + 2 O<sub>2</sub> → CO<sub>2</sub> + 2 H<sub>2</sub>O, the stoichiometric coefficient of CH<sub>4</sub> would be 1 and the stoichiometric coefficient of O<sub>2</sub> would be 2.
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| In more technically precise terms, the stoichiometric coefficient in a [[chemical reaction]] [[system]] of the ''i–th'' component is defined as
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| :<math>\nu_i = \frac{dN_i}{d\xi} \,</math>
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| or
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| :<math> dN_i = \nu_i d\xi \,</math> | |
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| where ''N''<sub>''i''</sub> is the number of [[molecule]]s of ''i'', and ξ is the progress [[variable (math)|variable]] or '''extent of reaction''' (Prigogine & Defay, p. 18; Prigogine, pp. 4–7; Guggenheim, p. 37 & 62).
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| <blockquote>
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| The '''extent of reaction ξ''' can be regarded as a real (or hypothetical) product, one molecule of which produced each time the reaction event occurs. It is the extensive quantity describing the progress of a chemical reaction equal to the number of chemical transformations, as indicated by the reaction equation on a molecular scale, divided by the Avogadro constant (in essence, it is the amount of chemical transformations). The change in the extent of reaction is given by dξ = d''n''<sub>B</sub>/''ν''<sub>B</sub>, where ''ν''<sub>B</sub> is the stoichiometric number of any reaction entity B (reactant or product) an d''n''<sub>B</sub> is the corresponding amount.<ref>[http://www.iupac.org/goldbook/E02283.pdf IUPAC Compendium of Chemical Terminology 2nd Edition (1997)]</ref>
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| </blockquote>
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| The stoichiometric coefficient ν<sub>''i''</sub> represents the degree to which a chemical species participates in a reaction. The convention is to assign negative coefficients to ''reactants'' (which are consumed) and positive ones to ''products''. However, any reaction may be viewed as "going" in the reverse direction, and all the coefficients then change sign (as does the [[Thermodynamic free energy|free energy]]). Whether a reaction actually ''will'' go in the arbitrarily selected forward direction or not depends on the amounts of the [[chemical substance|substances]] present at any given time, which determines the [[chemical kinetics|kinetics]] and [[thermodynamic equilibrium|thermodynamics]], i.e., whether [[chemical equilibrium|equilibrium]] lies to the ''right'' or the ''left''.
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| If one contemplates actual [[reaction mechanism]]s, stoichiometric coefficients will always be [[integer]]s, since elementary reactions always involve whole molecules. If one uses a composite representation of an "overall" reaction, some may be [[rational number|rational]] [[fraction (mathematics)|fractions]]. There are often chemical species present that do not participate in a reaction; their stoichiometric coefficients are therefore zero. Any chemical species that is regenerated, such as a [[catalyst]], also has a stoichiometric coefficient of zero.
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| The simplest possible case is an [[isomer]]ism
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| :<math> A \iff B </math> | |
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| in which ν<sub>''B''</sub> = 1 since one molecule of ''B'' is produced each time the reaction occurs, while ν<sub>''A''</sub> = −1 since one molecule of ''A'' is necessarily consumed. In any chemical reaction, not only is the total [[conservation of mass|mass conserved]] but also the numbers of [[atom]]s of each [[periodic table|kind]] are conserved, and this imposes corresponding constraints on possible values for the stoichiometric coefficients.
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| There are usually multiple reactions proceeding simultaneously in any [[nature|natural]] reaction system, including those in [[biology]]. Since any chemical component can participate in several reactions simultaneously, the stoichiometric coefficient of the ''i–th'' component in the ''k–th'' reaction is defined as
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| :<math>\nu_{ik} = \frac{\partial N_i}{\partial \xi_k} \,</math>
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| so that the total (differential) change in the amount of the ''i–th'' component is
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| :<math> dN_i = \sum_k \nu_{ik} d\xi_k. \,</math>
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| Extents of reaction provide the clearest and most explicit way of representing compositional change, although they are not yet widely used.
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| With complex reaction systems, it is often useful to consider both the representation of a reaction system in terms of the amounts of the chemicals present { ''N''<sub>''i''</sub> } ([[thermodynamic variable|state variables]]), and the representation in terms of the actual compositional [[Degrees of freedom (physics and chemistry)|degrees of freedom]], as expressed by the extents of reaction { ξ<sub>''k''</sub> }. The transformation from a [[vector space|vector]] expressing the extents to a vector expressing the amounts uses a rectangular [[matrix (mathematics)|matrix]] whose elements are the stoichiometric coefficients [ ν<sub>''i k''</sub> ].
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| The [[extreme value|maximum and minimum]] for any ξ<sub>''k''</sub> occur whenever the first of the reactants is depleted for the forward reaction; or the first of the "products" is depleted if the reaction as viewed as being pushed in the reverse direction. This is a purely [[kinematics|kinematic]] restriction on the reaction [[simplex]], a [[hyperplane]] in composition space, or ''N''‑space, whose [[dimension]]ality equals the number of ''[[linear independence|linearly-independent]]'' chemical reactions. This is necessarily less than the number of chemical components, since each reaction manifests a relation between at least two chemicals. The accessible region of the hyperplane depends on the amounts of each chemical species actually present, a contingent fact. Different such amounts can even generate different hyperplanes, all sharing the same algebraic stoichiometry.
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| In accord with the principles of [[chemical kinetics]] and [[thermodynamic equilibrium]], every chemical reaction is ''reversible'', at least to some degree, so that each equilibrium point must be an [[interior (topology)|interior point]] of the simplex. As a consequence, extrema for the ξ's will not occur unless an experimental system is prepared with zero initial amounts of some products.
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| The number of ''physically''-independent reactions can be even greater than the number of chemical components, and depends on the various reaction mechanisms. For example, there may be two (or more) reaction ''paths'' for the isomerism above. The reaction may occur by itself, but faster and with different intermediates, in the presence of a catalyst.
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| The (dimensionless) "units" may be taken to be [[molecule]]s or [[mole (unit)|moles]]. Moles are most commonly used, but it is more suggestive to picture incremental chemical reactions in terms of molecules. The ''N'''s and ξ's are reduced to molar units by dividing by [[Avogadro's number]]. While dimensional [[mass]] units may be used, the comments about integers are then no longer applicable.
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| ==Stoichiometry matrix==
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| In complex reactions, stoichiometries are often represented in a more compact form called the stoichiometry matrix. The stoichiometry matrix is denoted by the symbol, <math>\mathbf{N}</math>.
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| If a reaction network has <math> \mathit{n} </math> reactions and <math> \mathit{m} </math> participating molecular species then the stoichiometry matrix will have corresponding <math> \mathit{m} </math> rows and <math> \mathit{n} </math> columns.
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| For example, consider the system of reactions shown below:
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| :S<sub>1</sub> → S<sub>2</sub>
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| :5S<sub>3</sub> + S<sub>2</sub> → 4S<sub>3</sub> + 2S<sub>2</sub>
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| :S<sub>3</sub> → S<sub>4</sub>
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| :S<sub>4</sub> → S<sub>5</sub>.
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| This systems comprises four reactions and five different molecular species. The stoichiometry matrix for this system can be written as:
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| : <math>
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| \mathbf{N} = \begin{bmatrix}
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| -1 & 0 & 0 & 0 \\ | |
| 1 & 1 & 0 & 0 \\
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| 0 & -1 & -1 & 0 \\
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| 0 & 0 & 1 & -1 \\ | |
| 0 & 0 & 0 & 1 \\
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| \end{bmatrix}
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| </math>
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| where the rows correspond to S<sub>1</sub>, S<sub>2</sub>, S<sub>3</sub>, S<sub>4</sub> and S<sub>5</sub>, respectively. Note that the process of converting a reaction scheme into a stoichiometry matrix can be a lossy transformation, for example, the stoichiometries in the second reaction simplify when included in the matrix. This means that it is not always possible to recover the original reaction scheme from a stoichiometry matrix.
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| Often the stoichiometry matrix is combined with the rate vector, v to form a compact equation describing the rates of change of the molecular species:
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| : <math>
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| \frac{d\mathbf{S}}{dt} = \mathbf{N} \cdot \mathbf{v}.
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| </math>
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| ==Gas stoichiometry==
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| ''Gas stoichiometry'' is the quantitative relationship (ratio) between reactants and products in a [[chemical reaction]] with reactions that produce [[gases]]. Gas stoichiometry applies when the gases produced are assumed to be [[ideal gas|ideal]], and the temperature, pressure, and volume of the gases are all known. The ideal gas law is used for these calculations. Often, but not always, the [[standard temperature and pressure]] (STP) are taken as 0 °C and 1 bar and used as the conditions for gas stoichiometric calculations.
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| Gas stoichiometry calculations solve for the unknown [[volume]] or [[mass]] of a gaseous product or reactant. For example, if we wanted to calculate the volume of gaseous NO<sub>2</sub> produced from the combustion of 100 g of NH<sub>3</sub>, by the reaction:
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| :4NH<sub>3</sub> (g) + 7O<sub>2</sub> (g) → 4NO<sub>2</sub> (g) + 6H<sub>2</sub>O (l) | |
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| we would carry out the following calculations:
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| :<math> 100 \ \mbox{g}\,NH_3 \cdot \frac{1 \ \mbox{mol}\,NH_3}{17.034 \ \mbox{g}\,NH_3} = 5.871 \ \mbox{mol}\,NH_3\ </math>
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| There is a 1:1 molar ratio of NH<sub>3</sub> to NO<sub>2</sub> in the above balanced combustion reaction, so 5.871 mol of NO<sub>2</sub> will be formed. We will employ the [[ideal gas law]] to solve for the volume at 0 °C (273.15 K) and 1 atmosphere using the [[gas constant|gas law constant]] of R = 0.08206 L · atm · K<sup>−1</sup> · mol<sup>−1</sup> :
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| :{| border="0" cellpadding="2"
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| |-
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| !align=right|<math>PV</math>
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| |align=left|<math>= nRT</math>
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| |-
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| !align=right|<math>V</math>
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| |align=left|<math>= \frac{nRT}{P} = \frac{5.871 \cdot 0.08206 \cdot 273.15}{1} = 131.597 \ \mbox{L}\,NO_2</math>
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| |}
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| Gas stoichiometry often involves having to know the [[molar mass]] of a gas, given the [[density]] of that gas. The ideal gas law can be re-arranged to obtain a relation between the [[density]] and the [[molar mass]] of an ideal gas:
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| :<math>\rho = \frac{m}{V}</math> and <math>n = \frac{m}{M}</math>
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| and thus:
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| :<math>\rho = \frac {M P}{R\,T}</math>
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| {| border="0" cellpadding="2"
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| |-
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| |align=right|where:
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| |align=left|
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| |-
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| !align=right|<math>P</math>
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| |align=left|= absolute gas [[pressure]]
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| |-
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| !align=right|<math>V</math>
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| |align=left|= gas [[volume]]
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| |-
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| !align=right|<math>n</math>
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| |align=left|= number of [[mole (unit)|moles]]
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| |-
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| !align=right|<math>R</math>
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| |align=left|= universal ideal gas law constant
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| |-
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| !align=right|<math>T</math>
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| |align=left|= absolute gas [[temperature]]
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| |-
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| !align=right|<math>\rho</math>
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| |align=left|= gas density at <math>T</math> and <math>P</math>
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| |-
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| !align=right|<math>m</math>
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| |align=left|= mass of gas
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| |-
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| !align=right|<math>M</math>
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| |align=left|= molar mass of gas
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| |}
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| ==Stoichiometric air-to-fuel ratios of common fuels==
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| {{see also|Air–fuel ratio|Combustion}}
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| In the [[combustion]] reaction, oxygen reacts with the fuel, and the point where exactly all oxygen is consumed and all fuel burned is defined as the stoichiometric point. With more oxygen (overstoichiometric combustion), some of it stays unreacted. Likewise, if the combustion is incomplete due to lack of sufficient oxygen, fuel remains unreacted. (Unreacted fuel may also remain because of slow combustion or insufficient mixing of fuel and oxygen - this is not due to stoichiometry.) Different hydrocarbon fuels have a different contents of carbon, hydrogen and other elements, thus their stoichiometry varies.
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| {| class="wikitable"
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| |-
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| ! Fuel
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| ! Ratio by mass <ref>John B. Heywood: "Internal Combustion Engine Fundamentals page 915", 1988</ref>
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| ! Ratio by volume <ref>North American Mfg. Co.: "North American Combustion Handbook", 1952</ref>
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| ! Percent fuel by mass
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| |-
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| | Gasoline
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| | 14.7 : 1
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| | —
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| | 6.8%
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| |-
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| | Natural gas
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| | 17.2 : 1
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| | 9.7 : 1
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| | 5.8%
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| |-
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| | Propane (LP)
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| | 15.67 : 1
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| | 23.9 : 1
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| | 6.45%
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| |-
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| | Ethanol
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| | 9 : 1
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| | —
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| | 11.1%
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| |-
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| | Methanol
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| | 6.47 : 1
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| | —
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| | 15.6%
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| |-
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| | Hydrogen
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| | 34.3 : 1
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| | 2.39 : 1
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| | 2.9%
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| |-
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| | Diesel
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| | 14.5 : 1
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| | —
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| | 6.8%
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| |}
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| Gasoline engines can run at stoichiometric air-to-fuel ratio, because gasoline is quite volatile and is mixed (sprayed or carburetted) with the air prior to ignition. Diesel engines, in contrast, run lean, with more air available than simple stoichiometry would require. Diesel fuel is less volatile and is effectively burned as it is injected, leaving less time for evaporation and mixing. Thus, it would form soot (black smoke) at stoichiometric ratio.
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| ==Stoichiometric reactants and catalytic reactants==
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| A '''stoichiometric reactant''' is a reactant that is consumed in a reaction, as opposed to a [[catalysis|catalytic reactant]], which is not consumed in the overall reaction because it reacts in one step and is regenerated in another step.
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| ==References==
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| {{Reflist}}
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| *Zumdahl, Steven S. ''Chemical Principles''. Houghton Mifflin, New York, 2005, pp 148–150.
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| * Internal Combustion Engine Fundamentals, John B. Heywood
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| ==External links==
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| * [http://www.tech.plym.ac.uk/sme/ther305-web/Combust1.PDF Engine Combustion primer] from the University of Plymouth
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| * [http://www.chemcollective.org/tutorials.php Free Stoichiometry Tutorials] from Carnegie Mellon's ChemCollective
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| * [http://chemistry-in-excel.jimdo.com Stoichiometry Add-In for Microsoft Excel] for calculation of molecular weights, reaction coëfficients and stoichiometry.
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| * [http://www.thermobook.net/stoichiometry/ Reaction Stoichiometry Calculator] a comprehensive free online reaction stoichiometry calculator.
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| {{Library resources box
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| |by=no
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| |onlinebooks=no
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| |others=no
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| |about=yes
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| |label=Stoichiometry}}
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| [[Category:Stoichiometry| ]]
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| [[Category:Chemical engineering]]
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